A053026 Maximum power of 2 arising when A000005 is applied repeatedly to n!.
1, 2, 4, 8, 16, 8, 4, 4, 4, 16, 8, 8, 8, 8, 8, 2, 8, 4, 8, 4, 4, 4, 4, 4, 4, 4, 16, 16, 4, 16, 8, 16, 4, 4, 16, 4, 4, 2, 4, 16, 8, 16, 16, 4, 8, 8, 4, 8, 8, 16, 8, 8, 32, 32, 4, 32, 4, 4, 8, 4, 2, 32, 2, 8, 4, 8, 4, 8, 8, 8, 8, 2, 8, 8, 8, 32, 32, 8, 4, 8, 8, 4, 8, 8, 8, 8, 8, 32, 8, 8, 2, 4, 2, 4, 8
Offset: 1
Keywords
Examples
For n = 53, the iterations are {53!, 16174080000, 840, 32, 6, 4, 3, 2}, so a(53) = 32. For n = 130, the iterations are {130!, 287298761874053529600, 38016, 64, 7, 2}, so a(130) = 64. For n = 563, the iterations are {563!, 2875041108020454013464609906430286933482949481627276804096000000000, 77051520, 512, 10, 4, 3, 2}, so a(563) = 512.
Programs
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Mathematica
Join[{1,2},Table[SelectFirst[Rest[NestWhileList[DivisorSigma[0,#]&,n!,#>2&]],IntegerQ[Log[2,#]]&],{n,3,100}]] (* Harvey P. Dale, Jul 02 2018 *)
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PARI
a(n) = {my(m = n!); while(1 << valuation(m, 2) != m, m = numdiv(m)); m;} \\ Amiram Eldar, Feb 04 2025
Formula
a(n) = 2^A380802(n). - Amiram Eldar, Feb 04 2025
Comments